~~簡易不等式(1)~~

因為x,y 係實數, 所以會 x > y or x < y...(至於x=y之後再講...)
因為
(x + y) / (1+x+y) 及 x / (1+x) + y / (1+y) 係對稱,所以...
WLOG, SAY y > x
=> 1+y > 1+x
=>(1+y) + (x+y)(1+y) > 1+y > 1+x (since x,y>0)
=>(1+x+y)(1+y) > 1+x
Since x,y>0
1/(1+x) > 1/[(1+x+y)(1+y)]
(-x) / (1+x) < (-x) / [(1+x+y)(1+y)]
[1-(1+x)] / (1+x) < [(1+y)-(1+x+y)] / [(1+x+y)(1+y)]
1 / (1+x) -1 < 1 / (1+x+y) - 1 / (1+y)
1 / (1+x) + 1 / (1+y) -1 < 1 / (1+x+y)
-[1 / (1+x) + 1 / (1+y) -1] > -[1 / (1+x+y)]
1-[1 / (1+x)+1 / (1+y) -1] > 1-[1 / (1+x+y)]
[1-1 / (1+x)]+[1-1 / (1+y)] > 1- 1 / (x+y+1)
x / (1+x) + y / (1+y) > (x+y) / (x+y+1)
所以當x,y>0的時候(x不等於y),(x + y) / (1+x+y) 小過 x / (1+x) + y / (1+y)
好la~...when x=y,
(x + y) / (1+x+y) =(2x) / (2x+1)
x/(1+x)+y/(1+y) = (2x)/(1+x)
Since x>0,
x > 0
2x > x
1+2x > x+1
1 / (1+2x) < 1/ (x+1)
(2x) / (1+2x) < (2x) / (1+x)
=>(x + y) / (1+x+y) < x/(1+x)+y/(1+y)
As a result, 只要x,y>0, (x + y) / (1+x+y) 小過 x / (1+x) + y / (1+y)
但x,y兩個都等於0, (x + y) / (1+x+y) = x / (1+x) + y / (1+y)
x,y是但一個等於0,結果一樣...(x + y) / (1+x+y) = x / (1+x) + y / (1+y)

將兩式相減 看看是否正數還是負數
但過程繁複 要通幾次分母 然後化簡一堆分子

wkho28 :

Thank you !

Compare (x+y)/(1+x+y) and x/(1+x)+y/(1+y) :

[(x+y)/(1+x+y)]-[x/(1+x)+y/(1+y)]
=[(x+y)/(1+x+y)-x/(1+x)]-y/(1+y)
=[(x+y)(1+x)-x(1+x+y)]/[(1+x+y)(1+x)]-y/(1+y)
=(x+y+x^2+xy-x-x^2-xy)/[(1+x+y)(1+x)]-y/(1+y)
=y/[(1+x+y)(1+x)]-y/(1+y)
=y{1/[(1+x+y)(1+x)]-1/(1+y)}

2011-05-29 21:39:03 補充：
=y[(1+y)-(1+x+y)(1+x)]/[(1+x+y)(1+x)(1+y)]
=y(1+y-1-x-y-x-x^2-xy)/[(1+x+y)(1+x)(1+y)]
=y(-2x-x^2-xy)/[(1+x+y)(1+x)(1+y)]
=-xy(2+x+y)/[(1+x+y)(1+x)(1+y)]

Given that x>0 and y>0
therefore 2+x+y>0 , 1+x+y>0 , 1+x>0 and 1+y>0

2011-05-29 21:39:15 補充：
-xy(2+x+y)/[(1+x+y)(1+x)(1+y)]<0
i.e. [(x+y)/(1+x+y)]-[x/(1+x)+y/(1+y)]<0
(x+y)/(1+x+y)